3.8.7 Pengamiran, SPM Praktis (Kertas 2)

Soalan 8:
Rajah di bawah menunjukkan lengkung $y=\frac{4}{x^2}$ dan garis lurus y = mx + c. Garis lurus y = mx + c ialah tangen kepada lengkung pada (2, 1).
(a) Cari nilai m dan nilai c.

(b) Hitung luas kawasan berlorek.

(c) Diberi bahawa isi padu kisaran apabila rantau yang dibatasi oleh lengkung, paksi-x, garis lurus x = 2 dan x = h diputarkan melalui 360^o pada paksi-x ialah $\frac{38\pi }{81}{\text{ unit}}^3.$
Cari nilai h, dengan keadaan h > 2.


Penyelesaian:
(a)
$\begin{array}{l}y=\frac{4}{x^2}=4x^{-2}\\ \frac{dy}{dx}=-8x^{-3}\\ =-\frac{8}{x^3}\\ \text{At }x=2,\\ \frac{dy}{dx}=-\frac{8}{2^3}\\ =-1\\ \\ \text{Persamaan tangen:}\\ y-y_1=m\left(x-x_1\right)\\ y-1=-1\left(x-2\right)\\ y=-x+2+1\\ y=-x+3\\ \\ m=-1,c=3\end{array}$


(b)
$\begin{array}{l}\text{Pada paksi-}x\text{, }y=0\\ \text{Dari garis lurus }y=-x+3,x=3\\ \\ \text{Luas kawasan berlorek}\\ =\text{Luas bawah lengkung}-\text{Luas segi tiga}\\ ={\displaystyle {\int }_2^{4}ydx-}\frac{1}{2}\times 1\times 1\\ ={\displaystyle {\int }_2^4\left(4x^{-2}\right)dx}-\frac{1}{2}\\ ={\left[\frac{4x^{-1}}{-1}\right]}_2^4-\frac{1}{2}\\ ={\left[-\frac{4}{x}\right]}_2^4-\frac{1}{2}\\ =\left[-\frac{4}{4}-\left(-\frac{4}{2}\right)\right]-\frac{1}{2}\\ =\frac{1}{2}{\text{ unit}}^2\end{array}$


(c)
$\begin{array}{l}\text{Isipadu kisaran}=\frac{38\pi }{81}\\ \pi {\displaystyle {\int }_2^h{y}^{2}dx}=\frac{38\pi }{81}\\ {\displaystyle {\int }_2^h{\left(\frac{4}{x^2}\right)}^{2}d}x=\frac{38}{81}\\ {\displaystyle {\int }_2^h\left(\frac{16}{x^4}\right)dx}=\frac{38}{81}\\ {\displaystyle {\int }_2^h\left(16x^{-4}\right)dx}=\frac{38}{81}\\ {\left[\frac{16x^{-3}}{-3}\right]}_2^h=\frac{38}{81}\\ {\left[-\frac{16}{3x^3}\right]}_2^h=\frac{38}{81}\\ -\frac{16}{3h^3}-\left(-\frac{16}{3{\left(2\right)}^3}\right)=\frac{38}{81}\\ \frac{16}{3h^3}=\frac{16}{24}-\frac{38}{81}\\ \frac{16}{3h^3}=\frac{16}{81}\\ 3h^3=81\\ h^3=27\\ h=3\end{array}$